Which object is created during the formation of a star? a nebula a protostar a supergiant a supernova

Which statement best describes why a machine is useful?

sdas [7]

There are no options here. Could you provide the different answers?

7 0

3 years ago

Now find the electromotive force E2(t) induced across the entirety of solenoid 2 by the change in current in solenoid 1. Remembe

lions [1.4K]

Answer:

$E_{2} (t) = -\pi\mu_o} \rho^{2} n_{1}n_{2}L\frac{d }{dt}I_{1}(t)$

Explanation:

Consider two solenoids that are wound on a common cylinder as shown in fig. 1. Let the cylinder have radius 'ρ' and length 'L' .

No. of turns of solenoid 1 = n₁

No. of turns of solenoid 1 = n₂

Assume that length of solenoid is much longer than its radius, so that its field can be determined from Ampère's law throughout its entire length:

$\oint \overrightarrow {B}\overrightarrow {(r)}.\overrightarrow {dl}= \mu_{o}I$

We will consider the field that arises from solenoid 1, having n₁ turns per unit length. The magnetic field due to solenoid 1 passes through solenoid 2, which has n₂ turns per unit length.

Any change in magnetic flux from the field generated by solenoid 1 induces an EMF in solenoid 2 through Faraday's law of induction:

$\oint \overrightarrow {B}\overrightarrow {(r)}.\overrightarrow {dl}= -\frac{d}{dt} \phi _{M}(t)$

Consider B₁(t) magnetic feild generated in solenoid 1 due to current I₁(t)

Using:

$B_{1}(t) =\mu _{o} nI(t)\\$ --- (2)

Flux generated due to magnetic field B₁

$\phi _{1}(t) = \oint \overrightarrow {B_{1}}.dA\\$ ---(3)

area of solenoid = $A = \pi \rho^{2}$

substituting (2) in (3)

$\phi _{1}(t) = \mu_{o} \pi \rho^{2} n_{1}I_{1}(t)$ ----(4)

We have to find electromotive force E₂(t) induced across the entirety of solenoid 2 by the change in current in solenoid 1, i.e.

$E_{2} (t) = -n_{2}L\frac{d \phi_{1}}{dt}$ ---- (5)

substituting (4) in (5)

$E_{2} (t) = -n_{2}L\frac{d }{dt}(\mu_o} \pi \rho^{2} n_{1}I_{1}(t))\\E_{2} (t) = -\pi\mu_o} \rho^{2} n_{1}n_{2}L\frac{d }{dt}I_{1}(t)$

5 0

3 years ago

Which data set has the largest range? A. 55, 57, 59, 60, 61, 49, 48 B. 21, 25, 14, 16, 29, 22, 20 C. 12, 15, 16, 19, 18, 15, 27

Simora [160]

Data D has the largest range.

Data A: 61-48=13
Data B: 29-14=15
Data C:27-12=15
Data D:54-31=23

Therefore, Data D has the largest range.

3 0

3 years ago

A certain airplane has a speed of 290.0 km/h and is diving at an angle of 0 : 30.0" below the horizontal when the pilot releases

prisoha [69]

Answer:

10.03 s is the time the decoy is in the air

Explanation:

From the information given we know:

$\theta_{0}=-30$ because is measured clockwise from horizontal.

The initial speed of the decoy is the plane's speed at the moment of release $v_{0}=290 \:\frac{km}{h}$

We will need to work in meters and m/s, so $290 \:\frac{km}{h} \cdot \frac{1000 \:m}{km} \cdot \frac{1\:h}{3600\:s} = 80.56 \:\frac{m}{s}$

The horizontal motion of a projectile is given by:

$x-x_{0}=v_{0x}\cdot t$

because $v_{0x}=v_0cos(\theta_{0})$ , this becomes

$x-x_{0}=(v_0cos(\theta_{0}))\cdot t$

We have that $\Delta x= 700\:m$ , solving for <em>t, </em>we have

$t=\frac{\Delta x}{v_{0}cos\theta_{0}} \\t=\frac{700 \:m}{(80.56 \:\frac{m}{s} )\cdot cos(-30)} \\t= 10.03 \:s$

5 0

3 years ago

Review. A bead slides without friction around a loop-the loop (Fig. P8.5). The bead is released from rest at a height h=3.50R (b

oksian1 [2.3K]

The normal force on the bead at point A is 1 N.

According to the Law Of Conservation of Energy, the total energy of the bead at the topmost point of height of 3.50 R is the same as its total energy at point A.

The energy at a height of 3.50R is equal to its potential energy which is

U= mgh= mg x 3.5R= 3.5mgR .........(i)

The energy at point A is,

E= mg(2R)+ 1/2 mv^2 ...........(ii)

Now, equating (i) and (ii)

3.5mgR = 2mgR + 1/2 mv^2

3.5mgR - 2mgR = 1/2mv^2

1.5gR = 1/2 v^2

3gR = v^2

V = √3gR

At point A, three forces are there weight of the bead(mg), normal reaction(N), and centripetal force(mv^2/r).

Balancing these forces,

mg + N = mv^2/r

N= mv^2/r - mg

Putting the given values in the above equation, we get the normal reaction

N= 1 N

Thus, the normal force on the bead at point A is 1N.

To know more about "Conservation of Mechanical Energy", refer to the following link:

brainly.com/question/11264649?referrer=searchResults

#SPJ4

8 0

1 year ago